If X^Y denotes X raised to the power of Y, find out last two
digits of (2957^3661)+(3081^3643)
Options
o 42
o 38
o 98
o 22

• (2957)^3661=(2957)^3660 * (2957)^1=2957,
  so last two digits be 57.
  2nd (3081)^3643=(3081)^3640 * (3081)^3=>for last two digits it will be
  (81)^1 * (81)^2=81*6561=531441=>so last two digit will be 41............
  now sum 57+41=98.so 98 ans.
• 12 years agoHelpfull: Yes(31) No(6)
• (2957^4)^3660 *2957 +(3081)^3643
  last 2 digits of (57)^4 is 01
  so (01)^3660 yields last two digits as 01
  {since the unit's digit is always 1 and tens digit is obtained by 0*0=0}
  so 2957*01 gives last 2 digits 57
  similarly (81)^3643 gives ten's digit as the unit digit of 8*3 ie 4
  so 57 +41=98
  ans is 98
• 12 years agoHelpfull: Yes(15) No(15)
• 57 has repeating frequency of once every 4 numbers.
  81 has repeating frequency of once every 5 numbers.
  So ans=57^1+61(last two digit of 81^3)=98
• 12 years agoHelpfull: Yes(8) No(7)
• @Falgnuni-how is (2957)^3660 * (2957)^1=2957????????????????????
  
• 12 years agoHelpfull: Yes(6) No(1)
• unit digit of 8^3 is 2...not 4..how the 2nd digit of 81^3 is becomming 4..pls xplain...
• 12 years agoHelpfull: Yes(5) No(1)
• pls xplain ur ans...its not clear to me..
• 12 years agoHelpfull: Yes(5) No(0)
• For (2957^3661)
  Ist Do 7^1=7
  then 5*1=5
  5 is the last digit of 5*1
  total=57
  For:
  (3081^3643)
  1^3=1
  8*3=24
  take last digit 0f 24 i.e. 4
  total=41
  total sum of digits=41+57=98(ans)
  
  
• 10 years agoHelpfull: Yes(5) No(3)
• plz xplain ds clearly...its not clear
• 12 years agoHelpfull: Yes(4) No(0)
• (2957^3661)+(3081^3643)
  Unit DIGIT
  cycle of 7= 7,9,3,1
  3661%4=1 (% is mod operator)
  so (2957^3661) unit digit is 7.
  
  (3081^3643), unit digit of 1^n is 1
  FINAL UNIT DIGIT=8%10=8
  
  
  NOTE:- Tens digit of
  abcd^wxyz is c*z
  
  Tens digit of
  (2957^3661) =5x1=5
  Tens digit of
  (3081^3643)=8x3=24
  FINAL TENS DIGIT=24+5=29%10=9
  
  Ans 98
• 9 years agoHelpfull: Yes(3) No(0)